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In the 2nd graph, is there an asymptote?


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There is a vertical asymptote $x=2$ in both pictures. – Brian M. Scott Oct 1 '12 at 6:39
are you sure? but there is a solid black point at x=2 in the second graph o-o – John Lee Oct 1 '12 at 6:43
There's still an asymptote. Solid black point notwithstanding. Also, when someone goes to the trouble of giving you an answer (and a good answer too), don't you think it's a bit rude to ask if he's sure about it? – user22805 Oct 1 '12 at 6:46
The fact that the function has a defined value at $x=2$ doesn’t keep $x=2$ from being a vertical asymptote as $x\to 2^-$. Note that $x=2$ is not a vertical asymptote as $x\to 2^+$ in either picture, though for different reasons: in the first picture $x$ can’t approach $2$ from the right. – Brian M. Scott Oct 1 '12 at 6:47
It’s okay; I wasn’t offended. In fact it was helpful, since it pinned down what you were unclear about. – Brian M. Scott Oct 1 '12 at 6:48
up vote 0 down vote accepted

The definition of a vertical asymptote is important if you want to understand if something is an asymptote or not. From Calculus by Varberg, Purcell, and Rigdon:

The line $x = c$ is a vertical asymptote of the graph of $y = f(x)$ if any of the following four statements is true.

  1. $\lim\limits_{x \to c^+} f(x) = \infty$
  2. $\lim\limits_{x \to c^+} f(x) = -\infty$
  3. $\lim\limits_{x \to c^-} f(x) = \infty$
  4. $\lim\limits_{x \to c^-} f(x) = -\infty$

That's it. In both graphs you show, we have that statement 3, where $c = 2$, from the definition is true. Since one of those statements is true, the definition says that $x = 2$ is a vertical asymptote in both cases. Notice, the definition doesn't say anything about the value of the function at $x = c$, only the behavior of the graph as $x$ approaches $c$.

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